Test case of an elastic tetrahedron
-----------------------------------

Coordinates: 
1 0,0,0
2 1,0,0
3 0,1,0
4 0,0,1
Material:
Young = 1000
Poisson = 0.3
Loads:
P = -1 on nodes 2,3,4 in directions x,y,z respectively.
Boundary conditions:
u = 0 in plane y-z (displacement in x-dir.)
v = 0 in plane z-x (displacement in y-dir.)
w = 0 in plane x-y (displacement in z-dir.)

+--------------------------------------------------------------------------------------------+
| Node | Interpolation       | Nb. of    | Nb. of int. | Displacement   | Total | Rigid body |
|      |                     | dof/node  | pts         | u2=v3=w4 (*10) | dof   | no energy  |
+--------------------------------------------------------------------------------------------+
| 4    | Lagrange linear     | 3         |  5          | -0.024000      | 12    | Yes        |
|      |                     |           |  10         | -0.024000      |       |            |
+--------------------------------------------------------------------------------------------+
| 10   | Lagrange quadratic  | 3         |  5          | -0.199775      | 30    | Yes        |
|      | (on node per side)  |           |  10         | -0.199775      |       |            |
+--------------------------------------------------------------------------------------------+
| 16   | Lagrange cubic      | 3         |  14         | -0.209675      | 48    | No         |
|      | (two nodes per side)|           |  24         | -0.209675      |       |            |
+--------------------------------------------------------------------------------------------+
| 20   | Lagrange cubic      | 3         |  14         | -0.352515      | 60    | Yes        |
|      | (with faces bubbles)|           |  24         | -0.352515      |       |            |
+--------------------------------------------------------------------------------------------+
| 4    | Hermite I-C         | 12        |  14         | -0.291012      | 48    | Yes        |
+--------------------------------------------------------------------------------------------+
| 4    | Hermite II a        | 12        |  14         | -0.298709      | 48    | Yes        |
+--------------------------------------------------------------------------------------------+
| 4    | Hermite II b        | 12        |  14         | -0.248679      | 48    | Yes        |
+--------------------------------------------------------------------------------------------+

Solution II a : with imposed boundary conditions on displacements only.
         II b : with imposed boundary conditions on displacements and their derivatives.

Cubic 16 nodes should not be considered as a valid element since the rigid body
condition is not fulfilled.

Ratio strain energy per degree of freedom:

Linear        3*0.024 / 12 = 0.0060
Quadratic     3*0.200 / 30 = 0.0200
Cubic         3*0.352 / 60 = 0.0176
Hermite II a  3*0.299 / 48 = 0.0187

Linear interpolation is much too stiff. Other scheme give very similar results. 
In this elementary example, the quadratic element is the winner, with very simple 
and fast shape functions.

Ratio strain energy per free (excluding imposed) degrees of freedom:

Linear        3*0.024 / (12-9)  = 0.0240  (9  imposed)
Quadratic     3*0.200 / (30-16) = 0.0429  (16 imposed)
Cubic         3*0.352 / (60-30) = 0.0352  (30 imposed)
Hermite II a  3*0.299 / (48-9)  = 0.0230  (9  imposed)
Hermite II b  3*0.249 / (48-27) = 0.0356  (27 imposed including derivatives)

Again, quadratic element leads to the best results and requires only 5 integration points.
On the other hand, linear element requires one point only, compared to the 14 points
of cubic elements.