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Strain compatibility and displacement theory

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.

What are compatibility conditions of strain? Derive the compatibility equation for plane strain. A strain field is given by εx=2×10−6 y2\varepsilon_x = 2\times10^{-6}\,y^2, εy=3×10−6 x2\varepsilon_y = 3\times10^{-6}\,x^2, γxy=8×10−6 xy\gamma_{xy} = 8\times10^{-6}\,xy (x, y in mm). Check whether it is possible.

Compatibility conditions

The three strain components εx,εy,γxy\varepsilon_x, \varepsilon_y, \gamma_{xy} in a plane are obtained from only two displacement functions u(x,y)u(x,y) and v(x,y)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:

εx=∂u∂x,εy=∂v∂y,γxy=∂u∂y+∂v∂x\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:

∂2εx∂y2=∂3u∂x ∂y2∂2εy∂x2=∂3v∂y ∂x2∂2γxy∂x ∂y=∂3u∂x ∂y2+∂3v∂x2 ∂y\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):

∂2εx∂y2+∂2εy∂x2=∂2γxy∂x ∂y\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

∂2εx∂y2=2×2×10−6=4×10−6∂2εy∂x2=2×3×10−6=6×10−6LHS=10×10−6∂2γxy∂x ∂y=8×10−6=RHS\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 (10×10−6≠8×10−610\times10^{-6} \neq 8\times10^{-6}), so the strain field is not compatible: it cannot arise from any continuous displacement field. It would need γxy=10×10−6xy\gamma_{xy} = 10\times10^{-6}xy to be compatible.

Answer: The field is not possible, since 10×10−6≠8×10−610\times10^{-6}\neq 8\times10^{-6}.

✓ Answer checked by Claude Sonnet 5.5

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