# Introduction to 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*e1 + k*X*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
end do;
end do;

$gradu:={\begin{bmatrix}0&k&0\\k&0&0\\0&0&0\end{bmatrix}}$ 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.