# Introduction to Elasticity/Kinematics example 3

## Example 3

Given:

Unit square ${\displaystyle (X_{1},X_{2})\in [0,1]}$ with displacement fields :

1. ${\displaystyle \mathbf {u} =\kappa X_{2}{\widehat {\mathbf {e} }}_{1}+\kappa X_{1}{\widehat {\mathbf {e} }}_{2}}$.
2. ${\displaystyle \mathbf {u} =-\kappa X_{2}{\widehat {\mathbf {e} }}_{1}+\kappa X_{1}{\widehat {\mathbf {e} }}_{2}}$.
3. ${\displaystyle \mathbf {u} =\kappa X_{1}^{2}{\widehat {\mathbf {e} }}_{2}}$.

Sketch: Deformed configuration in ${\displaystyle x_{1},x_{2}}$ plane.

### Solution

The displacement ${\displaystyle \mathbf {u} =\mathbf {x} -\mathbf {X} }$. Hence, ${\displaystyle \mathbf {x} =\mathbf {u} +\mathbf {X} }$. In the reference configuration, ${\displaystyle \mathbf {u} =0}$ and ${\displaystyle \mathbf {x} =\mathbf {X} }$. Hence, in the ${\displaystyle (x_{1},x_{2})}$ plane, the initial square is the same shape as the unit square in the ${\displaystyle (X_{1},X_{2})}$ plane. We can use Maple to find out the values of ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ after the deformation ${\displaystyle \mathbf {u} }$.

  with(linalg):</code>
X := array(1..3): x := array(1..3): u = array(1..3):
e1 := array(1..3,[1,0,0]):
e2 := array(1..3,[0,1,0]): e3 = array(1..3,[0,0,1]):
ua := evalm(k*X[2]*e1 + k*X[1]*e2):
ub := evalm(-k*X[2]*e1 + k*X[1]*e2);
uc := evalm(k*X[1]^2*e2);

${\displaystyle {\mathit {ua}}:=\left[k{X_{2}},k{X_{1}},0\right]}$
${\displaystyle {\mathit {ub}}:=\left[-k{X_{2}},k{X_{1}},0\right]}$
${\displaystyle {\mathit {uc}}:=\left[0,k{X_{1}}^{2},0\right]}$
  xa := evalm(ua + X);
xb := evalm(ub + X);
xc := evalm(uc + X);</code>

${\displaystyle {\mathit {xa}}:=\left[k{X_{2}}+{X_{1}},k{X_{1}}+{X_{2}},{X_{3}}\right]}$
${\displaystyle {\mathit {xb}}:=\left[-k{X_{2}}+{X_{1}},k{X_{1}}+{X_{2}},{X_{3}}\right]}$
${\displaystyle {\mathit {xc}}:=\left[{X_{1}},k{X_{1}}^{2}+{X_{2}},{X_{3}}\right]}$

Plots of the deformed body are shown below

 Deformed shapes