Elasticity/Kinematics example 4

Example 4
 Given:

Displacement field $$\mathbf{u} = \kappa X_2 \widehat{\mathbf{e}}_{1} + \kappa X_1 \widehat{\mathbf{e}}_{2}$$.

 Find:


 * 1) The Lagrangian Green strain tensor $$\boldsymbol{E}\,$$.
 * 2) The infinitesimal strain tensor $$\boldsymbol{\varepsilon}\,$$.
 * 3) The infintesimal rotation tensor $$\boldsymbol{\omega}\,$$.
 * 4) The infinitesimal rotation vector $$\boldsymbol{\theta}\,$$.
 * 5) The exact longitudinal strain in the reference material direction $$\mathbf{e}_1\,$$.
 * 6) The approximate longitudinal strain in the direction $$\mathbf{e}_1\,$$ based on the infinitesimal strain tensor $$\boldsymbol{\varepsilon}\,$$.

Solution
The Maple output of the computations are shown below:

with(linalg): with(LinearAlgebra): X := array(1..3): x := array(1..3): e1 := array(1..3,[1,0,0]): e2 := array(1..3,[0,1,0]): e3 := array(1..3,[0,0,1]): u := evalm(k*X[2]*e1 + k*X[1]*e2);

u := \left[  \! k\,{X_{2}}, \,k\,{X_{1}}, \,0 \! \right] $$

x := evalm(u + X);

x := \left[  \! k\,{X_{2}} + {X_{1}}, \,k\,{X_{1}} + {X_{2}}, \, {X_{3}} \! \right] $$ F := linalg[matrix](3,3): for i from 1 to 3 do   for j from 1 to 3 do      F[i,j] := diff(x[i],X[j]); end do; end do; evalm(F);

F := \begin{bmatrix} 1 & k & 0 \\ k & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ Id := IdentityMatrix(3): C := evalm(transpose(F)&*F); E := evalm((1/2)*(C - Id));

C := \begin{bmatrix} 1 + k^{2} & 2\,k & 0 \\ 2\,k & 1 + k^{2} & 0 \\ 0 & 0 & 1   \end{bmatrix} $$

E := \begin{bmatrix} { \frac {k^{2}}{2}} & k & 0 \\ [2ex] k & { \frac {k^{2}}{2}} & 0 \\ [2ex] 0 & 0 & 0  \end{bmatrix} $$ gradu := linalg[matrix](3,3): for i from 1 to 3 do   for j from 1 to 3 do      gradu[i,j] := diff(u[i],X[j]); end do; end do; evalm(gradu);

gradu := \begin{bmatrix} 0 & k & 0 \\ k & 0 & 0 \\ 0 & 0 & 0    \end{bmatrix} $$ epsilon := evalm((1/2)*(gradu + transpose(gradu)));

\varepsilon := \begin{bmatrix} 0 & k & 0 \\ k & 0 & 0 \\ 0 & 0 & 0    \end{bmatrix} $$ omega := evalm((1/2)*(gradu - transpose(gradu)));

\omega := \begin{bmatrix} 0 & 0 & 0 \\      0 & 0 & 0 \\       0 & 0 & 0    \end{bmatrix} $$ stretch1 :=  sqrt(evalm(evalm(e1&*C)&*transpose(e1))[1,1]): longStrain1 := stretch1 - 1;

\mathit{stretch1} := \sqrt{1 + k^{2}} $$

\mathit{longStrain1} := \sqrt{1 + k^{2}} - 1 $$ approxLongStrain1 := evalm(evalm(e1&*epsilon)&*transpose(e1))[1,1];

\mathit{approxLongStrain1} := 0 $$

The geometrical difference between the large strain and small strain cases can be observed by looking at the figures from the previous examples.