Elasticity/Kinematics example 3

Example 3
 Given:

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


 * 1) $$\mathbf{u} = \kappa X_2 \widehat{\mathbf{e}}_{1} + \kappa X_1 \widehat{\mathbf{e}}_{2}$$.
 * 2) $$\mathbf{u} = -\kappa X_2 \widehat{\mathbf{e}}_{1} + \kappa X_1 \widehat{\mathbf{e}}_{2}$$.
 * 3) $$\mathbf{u} = \kappa X_1^2 \widehat{\mathbf{e}}_{2}$$.

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

Solution
The displacement $$\mathbf{u} = \mathbf{x} - \mathbf{X}$$. Hence, $$\mathbf{x} = \mathbf{u} + \mathbf{X}$$. In the reference configuration, $$\mathbf{u} = 0$$ and $$\mathbf{x} = \mathbf{X}$$. 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 $$\mathbf{u}$$.

with(linalg): 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);

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

\mathit{ub} := \left[    - k{X_{2}}, k{X_{1}}, 0   \right] $$

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

\mathit{xa} := \left[   k{X_{2}} + {X_{1}}, k{X_{1}} + {X_{2}}, {X_{3}}   \right] $$

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

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

Plots of the deformed body are shown below