# Why do we eliminate centre of gravity as degree of freedom in FEM

In the finite element lecture we learned, that it is impractical to use centre of gravity of a triangle as a degree of freedom. So we defined a subspace $P_3'$, where $P_2\subseteq P_3'\subseteq P_3$ $$P_3'=\{p\in P_3;\ p(a_{123})=-\frac{1}{6}\sum_{l=1,2,3}p(a_l)+\frac{1}{4}\sum_{l,m=1,2,3,\ l\neq m}p(a_{llm})\}$$ where $a_1$, $a_2$, $a_3$ form a triangle in 2 dimensions, $a_{llm}=\frac{1}{3}(2a_l+a_m)$.

What is the reason for this? Why don't we use the centre of gravity?

• What PDE are you trying to solve with this finite element space?
– Paul
Jan 3 '17 at 23:18
• no particular. is this convenient for some specific type of problems? Jan 3 '17 at 23:50
• Are you referring to the technique of "static condensation"? Jan 4 '17 at 1:47

No you can't use the center of an element as the center of gravity. The mass of the element at the center of gravity must be distributed to the corner nodes. $[ m]=r \int{ n^Tndv}$ where $[m]$ is the mass matrix, $r$ the density, and $n$ the shape function. $v$ is a volume or area depending whether it is a 3d or 2d problem.