Introduction to Elasticity/Kinematics example 3

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[edit] 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 x1,x2 plane.

[edit] 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 (x1,x2) plane, the initial square is the same shape as the unit square in the (X1,X2) plane. We can use Maple to find out the values of x1 and x2 after the deformation \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);
\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);</code>
\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

Deformed shapes
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