# Introduction to Elasticity/Kinematics example 4

## Example 4

Given:

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

Find:

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

### 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);

${\displaystyle u:=\left[\!k\,{X_{2}},\,k\,{X_{1}},\,0\!\right]}$
  x := evalm(u + X);

${\displaystyle x:=\left[\!k\,{X_{2}}+{X_{1}},\,k\,{X_{1}}+{X_{2}},\,{X_{3}}\!\right]}$
  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);

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

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

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

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

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

${\displaystyle {\mathit {stretch1}}:={\sqrt {1+k^{2}}}}$
${\displaystyle {\mathit {longStrain1}}:={\sqrt {1+k^{2}}}-1}$
  approxLongStrain1 := evalm(evalm(e1&*epsilon)&*transpose(e1))[1,1];

${\displaystyle {\mathit {approxLongStrain1}}:=0}$

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