A unit-square, unit-thickness bilinear quadrilateral is used in plane strain for an…

A unit-square, unit-thickness bilinear quadrilateral is used in plane strain for an isotropic material with shear modulus \(\mu=1\) and bulk modulus \(K=10^5\). For any displacement field, define \(e=\nabla\!\cdot u\), \(\varepsilon=\tfrac12(\nabla u+\nabla u^T)\), and \(\varepsilon_{\mathrm{dev}}=\varepsilon-\tfrac12 eI\). The element energy is \[ U=\int_0^1\!\int_0^1\left(\mu\,\varepsilon_{\mathrm{dev}}:\varepsilon_{\mathrm{dev}}+\frac K2 e^2\right)dx\,dy. \] The four nodes are \((0,0),(1,0),(1,1),(0,1)\), and their displacement values are taken from \(u(x,y)=(\alpha xy,0)\) with \(\alpha=0.02\); this field is represented exactly by the bilinear interpolation. Compute \(U\) when both terms are integrated by \(2\times2\) Gauss quadrature. Then compute the selectively reduced value when the deviatoric term still uses \(2\times2\) quadrature but the volumetric term uses one point at the center with unit-square weight \(1\). Report the volumetric and deviatoric contributions separately and determine the ratio of the two total energies.

Compute \(e\) first, then subtract \(eI/2\) from the two-dimensional strain tensor.