# 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, 2012 at 5:49
• Linear elasticity at the moment, but the question is about nonlinear elasticity in general. Jan 3, 2012 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". Jan 3, 2012 at 11:21
• Equations added. Jan 3, 2012 at 17:55
• With dPhi/dx, do you mean the gradient? Jan 4, 2012 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. Jan 4, 2012 at 21:03