--- title: "Strain compatibility and displacement theory" subject: "Mechanics of Solids" chapter: 3 key: "semester_5/mechanics_of_solids:compatibility-small-displacement-check" source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/compatibility-small-displacement-check site: BCT Engineering --- # Strain compatibility and displacement theory Mechanics of Solids · Chapter 3: Deformable Body and Strain Tensor - Asked in: year not given, 6 marks - Video: [Prob 4-44 Solids: Compatibility Equation](https://www.youtube.com/watch?v=MpnNr1gZXHc) — Dr Leonardo Bueno - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/compatibility-small-displacement-check - Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-3 ## Question What are compatibility conditions of strain? Derive the compatibility equation for plane strain. A strain field is given by $\varepsilon_x = 2\times10^{-6}\,y^2$, $\varepsilon_y = 3\times10^{-6}\,x^2$, $\gamma_{xy} = 8\times10^{-6}\,xy$ (x, y in mm). Check whether it is possible. ## Short answer For plane strain, compatibility condition is ∂²ε_x/∂y² + ∂²ε_y/∂x² = ∂²γ_xy/∂x∂y. With given field, LHS = 10×10⁻⁶, RHS = 8×10⁻⁶; not equal, so the strain field is not possible. ## Answer #### Compatibility conditions The three strain components $\varepsilon_x, \varepsilon_y, \gamma_{xy}$ in a plane are obtained from only two displacement functions $u(x,y)$ and $v(x,y)$. The strains are therefore not independent: they must satisfy a condition so that a **continuous single-valued** displacement field exists (the deformed body has no gaps or overlaps). This is the compatibility condition. #### Derivation From the small-displacement relations: $$ \varepsilon_x = \frac{\partial u}{\partial x},\quad \varepsilon_y = \frac{\partial v}{\partial y},\quad \gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} $$ Differentiate: $$ \begin{aligned} \frac{\partial^2 \varepsilon_x}{\partial y^2} &= \frac{\partial^3 u}{\partial x\,\partial y^2} \\ \frac{\partial^2 \varepsilon_y}{\partial x^2} &= \frac{\partial^3 v}{\partial y\,\partial x^2} \\ \frac{\partial^2 \gamma_{xy}}{\partial x\,\partial y} &= \frac{\partial^3 u}{\partial x\,\partial y^2} + \frac{\partial^3 v}{\partial x^2\,\partial y} \end{aligned} $$ Adding the first two and comparing with the third (the order of differentiation can be interchanged): $$ \frac{\partial^2 \varepsilon_x}{\partial y^2} + \frac{\partial^2 \varepsilon_y}{\partial x^2} = \frac{\partial^2 \gamma_{xy}}{\partial x\,\partial y} $$ In 3D there are six such equations. For a simply connected body, they are necessary and sufficient. #### Check of the given field $$ \begin{aligned} \frac{\partial^2 \varepsilon_x}{\partial y^2} &= 2\times 2\times10^{-6} = 4\times10^{-6} \\ \frac{\partial^2 \varepsilon_y}{\partial x^2} &= 2\times 3\times10^{-6} = 6\times10^{-6} \\ \text{LHS} &= 10\times10^{-6} \\ \frac{\partial^2 \gamma_{xy}}{\partial x\,\partial y} &= 8\times10^{-6} = \text{RHS} \end{aligned} $$ LHS $\neq$ RHS ($10\times10^{-6} \neq 8\times10^{-6}$), so the strain field is **not compatible**: it cannot arise from any continuous displacement field. It would need $\gamma_{xy} = 10\times10^{-6}xy$ to be compatible. **Answer:** The field is not possible, since $10\times10^{-6}\neq 8\times10^{-6}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why must strain components satisfy compatibility conditions? A. Because three strains come from only two displacement functions, so a continuous single-valued displacement must exist B. To make stress symmetric C. To ensure strains are small D. To satisfy equilibrium Answer: A. The strains are not independent; compatibility ensures no gaps or overlaps in the deformed body. (Key point: Meaning of compatibility) 2. Which is the plane compatibility equation? A. d2ex/dx2 + d2ey/dy2 = d2 gamma/dxdy B. d2ex/dy2 + d2ey/dx2 = d2 gamma_xy/dxdy C. d2ex/dy2 - d2ey/dx2 = gamma_xy D. ex + ey = gamma_xy Answer: B. Differentiating the strain-displacement relations and eliminating u and v gives this equation. (Key point: Plane compatibility equation) 3. For the given field, what are the two sides of the compatibility check? A. LHS = 4e-6, RHS = 8e-6 B. LHS = 6e-6, RHS = 8e-6 C. LHS = 10e-6, RHS = 8e-6 D. LHS = 10e-6, RHS = 10e-6 Answer: C. d2ex/dy2 = 4e-6 and d2ey/dx2 = 6e-6 give 10e-6, but d2 gamma/dxdy = 8e-6. (Key point: Check of given field) 4. What is the conclusion, and what would make the field compatible? A. Compatible; no change needed B. Not compatible; gamma_xy = 10e-6 xy would fix it C. Not compatible; gamma_xy = 6e-6 xy would fix it D. Compatible only for x = y Answer: B. 10e-6 does not equal 8e-6, so no continuous displacement exists; gamma_xy = 10e-6 xy would satisfy the equation. 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