# Are 8 Gauss points required for second order hexahedral finite elements?

Is it possible to get second order accuracy for hexahedral finite elements with fewer than 8 Gauss points without introducing unphysical modes? A single central Gauss point introduces an unphysical shearing mode, and the standard symmetric arrangement of 8 Gauss points is expensive compared to tetrahedral discretizations.

Edit: Someone asked for equations. The equations I'm interested in are nonlinear elasticity, either dynamic or quasistatic. The quasistatic equations are

$$\nabla \cdot P\left(\nabla \phi \right) = 0$$

where $\phi : \Omega \to \mathbf{R}^3$, $\Omega \subset \mathbf{R}^3$, and $P : \mathbf{R}^{3 \times 3} \to \mathbf{R}^{3 \times 3}$ is a hyperelastic first Piola-Kirchoff stress function. An simple example is compressible neo-Hookean, where $$P(F) = \mu (F - F^{-T}) +\lambda F^{-T} \log \det F$$

• What exactly are you simulating? – Dan Jan 3 '12 at 5:49
• Linear elasticity at the moment, but the question is about nonlinear elasticity in general. – Geoffrey Irving Jan 3 '12 at 7:41
• You should probably include the equations you're interested in, since the definition of "unphysical" depends on them. Or at least define precisely the space of functions that are "physical". – David Ketcheson Jan 3 '12 at 11:21
• Equations added. – Geoffrey Irving Jan 3 '12 at 17:55
• With dPhi/dx, do you mean the gradient? – Wolfgang Bangerth Jan 4 '12 at 1:34

• To expand on Jed's point: the reason the above "obvious" argument is false is that each quadrature point sees a $3 \times 3$ matrix. For tetrahedra, that covers all motions of the vertices excluding uniform translation, which doesn't affect energy or forces, so one quadrature point is sufficient for first order accuracy. – Geoffrey Irving Jan 4 '12 at 21:03