An Airy stress function in a 2D plane stress elasticity problem without body forces is given by $\phi(x, y) = c_1 x^4 + c_2 x^2 y^2 + c_3 y^4$. For this function to represent a mathematically valid elasticity solution, it must satisfy the biharmonic compatibility equation $\nabla^4 \phi = 0$. The relationship that the constants $c_1, c_2,$ and $c_3$ must satisfy is:

(A) $3 c_1 + c_2 + 3 c_3 = 0$

(B) $c_1 + 2 c_2 + c_3 = 0$

(C) $6 c_1 + c_2 + 6 c_3 = 0$

(D) $c_1 + c_2 + c_3 = 0$

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### 1. Strategy & Theoretical Principles In 2D isotropic linear elasticity, an Airy stress function automatically satisfies equilibrium when stresses are defined as: $$\sigma_{xx} = \frac{\partial^2 \phi}{\partial y^2}, \quad \sigma_{yy} = \frac{\partial^2 \phi}{\partial x^2}, \quad \tau_{xy} = -\frac{\partial^2 \phi}{\partial x \partial y}$$ For strain compatibility to be satisfied, $\phi$ must satisfy the biharmonic equation: $$\nabla^4 \phi = \frac{\partial^4 \phi}{\partial x^4} + 2 \frac{\partial^4 \phi}{\partial x^2 \partial y^2} + \frac{\partial^4 \phi}{\partial y^4} = 0$$ ### 2. Analytical Formulation Compute the fourth partial derivatives of $\phi = c_1 x^4 + c_2 x^2 y^2 + c_3 y^4$: 1. Fourth derivative with respect to $x$: $$\frac{\partial \phi}{\partial x} = 4 c_1 x^3 + 2 c_2 x y^2$$ $$\frac{\partial^2 \phi}{\partial x^2} = 12 c_1 x^2 + 2 c_2 y^2$$ $$\frac{\partial^3 \phi}{\partial x^3} = 24 c_1 x$$ $$\frac{\partial^4 \phi}{\partial x^4} = 24 c_1$$ 2. Mixed fourth derivative: $$\frac{\partial^2 \phi}{\partial y^2} = 2 c_2 x^2 + 12 c_3 y^2$$ $$\frac{\partial^3 \phi}{\partial x \partial y^2} = 4 c_2 x$$ $$\frac{\partial^4 \phi}{\partial x^2 \partial y^2} = 4 c_2$$ 3. Fourth derivative with respect to $y$: $$\frac{\partial^4 \phi}{\partial y^4} = 24 c_3$$ ### 3. Step-by-Step Calculation Substitute these into the biharmonic equation: $$\nabla^4 \phi = 24 c_1 + 2(4 c_2) + 24 c_3 = 0$$ $$24 c_1 + 8 c_2 + 24 c_3 = 0$$ Dividing through by 8: $$3 c_1 + c_2 + 3 c_3 = 0$$ ### 4. Distractor Forensics - **Option A ($3 c_1 + c_2 + 3 c_3 = 0$):** Exactly satisfies $\nabla^4 \phi = 0$. - **Option B ($c_1 + 2 c_2 + c_3 = 0$):** Forgets the factorial factor 24 ($4!$) of $x^4$ and $y^4$. - **Option C:** Divided incorrectly. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Notice the factor of 2 multiplying the mixed derivative $\frac{\partial^4 \phi}{\partial x^2 \partial y^2}$ in $\nabla^4 = (\nabla^2)^2$. - **Standard Reference:** Timoshenko, S.P. & Goodier, J.N., *Theory of Elasticity*, 3rd Edition, McGraw-Hill, Chapter 2.

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