4. Treatment of 2DA and 3DA cases

In these two cases, we were able to find a node from the list of knots to be rigidified that carried all the degrees of freedom of the solid. Let \(A\) be this node, then:

• In 2D: \(\mathit{DX}\), \(\mathit{DY}\), \(\mathit{DRZ}\);

• In 3D: \(\mathit{DX}\), \(\mathit{DY}\), \(\mathit{DZ}\), \(\mathit{DRX}\),, \(\mathit{DRY}\), \(\mathit{DRZ}\)

Let it be a \(M\) node from the list of nodes to be rigidified of any one. In theory of small displacements, the motion of a solid body is expressed by:

(4.1)\[ {U} _ {M}\ mathrm {=} {=} {U} _ {A} +\ theta\ Lambda\ text {AM}\]

4.1. 2DA case

We write linear relationships:

(4.2)\[\begin{split} \ begin {array} {}\ forall M\ne A:\left\ {\ begin {array} {\ begin {array} {} {}\ text {DX} (M) -\ text {DX} (A) +y\ text {DRZ} (A) =0\\\ text {DY} (A) =0\\\ text {} (A) =0\\\ text {} (A) =0\\\ text {DY} (A) =0\\\ text {DY} (A) =0\\\ text {DY} (A) =0\\\ text {DY} (A) =0\\\ text {DY} (A) =0\\\ text {DY} (A) =0\\\ text {DY} (A) =0\\\ text {DY} (A m {with}\ mathrm {AM} =(\ begin {array} {} {} x\\ y\ end {array})\\ +\ text {si} M\ text {door}\ text {DRZ}:\ text {}:\ text {}:\ text {DRZ} (M) (M) -\ text {} (M) -\ text {DRZ} (A) =0\ end {array}:\ text {}:\ text {} (M) -\ text {} (M) -\ text {} (A) =0\ end {array} DRZ\end{split}\]

4.2. 3DA case

\(\begin{array}{}\forall M\ne A:\left\{\begin{array}{}\text{DX}(M)-\text{DX}(A)-\text{DRY}(A)\text{.}z+\text{DRZ}(A)\text{.}y=0\\ \text{DY}(M)-\text{DY}(A)-\text{DRZ}(A)\text{.}x+\text{DRX}(A)\text{.}z=0\\ \text{DZ}(M)-\text{DZ}(A)-\text{DRX}(A)\text{.}y+\text{DRY}(A)\text{.}x=0\end{array}\right\}\\ \\ +\text{si}M\text{porte}\text{DRX},\text{DRY},\text{DRZ}:\left\{\begin{array}{}\text{DRX}(M)-\text{DRX}(A)=0\\ \text{DRY}(M)-\text{DRY}(A)=0\\ \text{DRZ}(M)-\text{DRZ}(A)=0\end{array}\right\}\end{array}\)