Elasticity/Kinematics example 3

Example 3

Given:

Unit square $(X_1,X_2)\in [0,1]$ with displacement fields :

1. $𝐮=\kappa X_2{\widehat{𝐞}}_1+\kappa X_1{\widehat{𝐞}}_2$.
2. $𝐮=-\kappa X_2{\widehat{𝐞}}_1+\kappa X_1{\widehat{𝐞}}_2$.
3. $𝐮=\kappa X_1^2{\widehat{𝐞}}_2$.

Sketch: Deformed configuration in $x_1,x_2$ plane.

Solution

The displacement $𝐮=𝐱-𝐗$. Hence, $𝐱=𝐮+𝐗$. In the reference configuration, $𝐮=0$ and $𝐱=𝐗$. Hence, in the $(x_1,x_2)$ plane, the initial square is the same shape as the unit square in the $(X_1,X_2)$ plane. We can use Maple to find out the values of $x_1$ and $x_2$ after the deformation $𝐮$.

  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);

$𝑢𝑎:=[k{X}_2,k{X}_1,0]$

$𝑢𝑏:=[-k{X}_2,k{X}_1,0]$

$𝑢𝑐:=[0,k{X_1}^2,0]$

  xa := evalm(ua + X);
  xb := evalm(ub + X);
  xc := evalm(uc + X);</code>

$𝑥𝑎:=[k{X}_2+X_1,k{X}_1+X_2,X_3]$

$𝑥𝑏:=[-k{X}_2+X_1,k{X}_1+X_2,X_3]$

$𝑥𝑐:=[X_1,k{X_1}^2+X_2,X_3]$

Plots of the deformed body are shown below

File:Kinematics ex3 fig.png Deformed shapes