Past exam of the elasticity and plasticity course of the University of Oxford / 2009 / 2 / b

The Airy stress function determines the in-plane stress through its second derivatives. If two functions give the same stress, their difference $B$ has $B_{xx}=B_{xy}=B_{yy}=0$. On a connected domain this implies that its gradient is constant. Thereforeis exactly the local affine gauge freedom. It also preserves biharmonicity. An imposed decay condition for $A$ can fix this freedom; adding an arbitrary biharmonic function would usually change the physical stress and is not a gauge transformation.

The source uses the forward Fourier phase $e^{ikx}$, so $\widehat{{\partial }_{x}A}=-ik\hat{A}$. For a nonzero wavenumber put $q=|k|$. Biharmonicity becomes $({\partial }_y^2-q^2{)}^2\hat{A}=0$. The decaying solutions are $(C+Dy)e^{-qy}$. The normal-stress condition gives $-k^2\hat{A}(k,0)=-\hat{P}(k)$, hence $C=\hat{P}/k^2$. The shear condition gives ${\hat{A}}_y(k,0)=0$, so $D=qC$. ConsequentlyIts second derivatives are $\widehat{A_{xx}}=-\hat{P}(1+qy)e^{-qy}$ and $\widehat{A_{yy}}=\hat{P}(qy-1)e^{-qy}$. The given strain/stress inversion therefore yieldsIntegrating in $y$ with the stated decaying-mode condition givesThis is the Fourier solution of a loaded elastic half-space.

The zero wavenumber needs separate treatment: these displayed divisions by $k^2$ and $|k|$ are not formulas at $k=0$. In particular a nonzero resultant $\hat{P}(0)=\int Pdx$ is incompatible with an ordinary decaying zero-mode displacement of the stipulated kind; two-dimensional net loading produces a logarithmic displacement reference issue. Balanced loads or an appropriate distributional/reference-displacement formulation are needed for a global transform solution. The nonzero modes and the displacement-gradient kernel below remain the correct formal calculation; the singular zero mode must not be silently assigned a finite value.

Multiplication of the preceding displacement formula by $-ik$ givesFor the supplied kernels, $\widehat{f_{1,x}}=-ik{e}^{-qy}$ and ${\hat{f}}_2=i(k/q)e^{-qy}$. The convolution theorem therefore givesSince $f_{1,x}(s,y)=-2ys/[\pi (s^2+y^2{)}^2]$ and $f_2(s,y)=s/[\pi (s^2+y^2)]$, combine the two fractions and divide by two:The sign is tied to the positive forward Fourier phase used here. This is a displacement derivative, so an additive displacement reference drops out. At $y>0$ the kernel is regular near $s=0$; the usual decay/integrability of the applied load controls the tails. A boundary-limit singular integral, if wanted at $y=0$, would require its own principal-value interpretation.