Elasticity/Kinematics example 2

Example 2

Given: A body occupies the unit cube $X_i\in [0,1]$ in the reference configuration. The mapping between the current and the reference configuration given by $x_1=X_1+\kappa X_2$, $x_2=X_2$, $x_3=X_3$.

Find:

1. Sketch current configuration.
2. Show that motion is isochoric.
3. Find stretches in the directions ${\widehat{𝐞}}_1$, ${\widehat{𝐞}}_2$, $(1/\sqrt{2})({\widehat{𝐞}}_1+{\widehat{𝐞}}_2)$, and $(1/\sqrt{2})({\widehat{𝐞}}_1-{\widehat{𝐞}}_2)$.
4. Find the orthogonal shear strain between the reference material directions ${\widehat{𝐞}}_1$ and ${\widehat{𝐞}}_2$. Also between directions $(1/\sqrt{2})({\widehat{𝐞}}_1+{\widehat{𝐞}}_2)$ and $(1/\sqrt{2})({\widehat{𝐞}}_1-{\widehat{𝐞}}_2)$.
5. Find principal stretches and principal directions of stretch ($\kappa =0.4$).

Solution:

File:Kinematics ex2 fig1.png Deformed shape.

• All parallelograms that have the equal heights and the same base have equal areas. Hence, there is no volume change in this deformation. Hence isochoric.

• Stretches in the material direction $\widehat{𝐆}$ are given by

$\lambda (\widehat{𝐆})=\sqrt{\widehat{𝐆}:𝑪:\widehat{𝐆}}$

where $𝑪$ is the Cauchy-Green deformation tensor

$𝑪={𝑭}^T\bullet 𝑭$

We will use Maple to calculate the stretches in the four directions.

      with(linalg):
      x := array(1..3): X := array(1..3): 
      x[1] := X[1] + k*X[2]: x[2] := X[2]: x[3] := X[3]:
      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);
      C := evalm(transpose(F)&*F);
      e1 := linalg[matrix](1,3,[1,0,0]): 
      e2 := linalg[matrix](1,3,[0,1,0]):
      e1pe2 := evalm((e1 + e2)/sqrt(2)): 
      e1me2 := evalm((e1-e2)/sqrt(2)):''
      lambda[1] := sqrt(evalm(evalm(e1&*C)&*transpose(e1))[1,1]);
      lambda[2] := sqrt(evalm(evalm(e2&*C)&*transpose(e2))[1,1]);
      lambda[3] := simplify(sqrt(evalm(evalm(e1pe2&*C)&*transpose(e1pe2))[1,1]));
      lambda[4] := simplify(sqrt(evalm(evalm(e1me2&*C)&*transpose(e1me2))[1,1]));

$F:=\left[\begin{array}{ccc}1& k& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$C:=\left[\begin{array}{ccc}1& k& 0\\ k& k^2+1& 0\\ 0& 0& 1\end{array}\right]$

${\lambda }_1:=1$

${\lambda }_2:=\sqrt{k^2+1}$

${\lambda }_3:=\frac{\sqrt{4+4k+2k^2}}{2}$

${\lambda }_4:=\frac{\sqrt{4-4k+2k^2}}{2}$

The following figure shows that the calculated stretches are correct.

File:Kinematics ex2 fig2.png Stretches

• The orthogonal shear strain between two orthogonal units vectors ${\widehat{𝐆}}_1$ and ${\widehat{𝐆}}_2$ in the reference material co-ordinate system is given by

$\gamma ({\widehat{𝐆}}_1,{\widehat{𝐆}}_2)={\sin }^{-1}\left(\frac{{\widehat{𝐆}}_1:𝐂:{\widehat{𝐆}}_2}{\lambda ({\widehat{𝐆}}_1)\lambda ({\widehat{𝐆}}_2)}\right)$

Once again, we will use Maple to calculate these strains. The steps are the same upto the calculation of the stretches.

      numer1 := evalm(evalm(e1&*C)&*transpose(e2))[1,1]:
      denom1 := lambda[1]*lambda[2]:
      ratio1 := numer1/denom1:
      gam1 := arcsin(ratio1);
      numer2 := simplify(evalm(evalm(e1pe2&*C)&*transpose(e1me2))[1,1]):
      denom2 := lambda[3]*lambda[4]:
      ratio2 := numer2/denom2:
      gam2 := arcsin(ratio2);

${\gamma }_1:={\sin }^{-1}(\frac{k}{\sqrt{k^2+1}})$

${\gamma }_2:=-{\sin }^{-1}(\frac{2k^2}{\sqrt{4+4k+2k^2}\sqrt{4-4k+2k^2}})$

The following figure shows that the calculated orthogonal shear strains are correct.

File:Kinematics ex2 fig3.png Orthogonal shear strains

The value of ${\gamma }_1$ can easily be verified using the definition of $\sin \gamma$. For the verification of the value of ${\gamma }_2$, the cosine law

$\cos (A)=\frac{b^2+c^2-a^2}{2bc}$

has to be used. Note that the actual length of the sides $b$ and $c$ of the triangle is obtained after multiplying the values shown in the figure by the length of the diagonal ($\sqrt{2}$). Hence, the calculated values are correct.

      b := sqrt(2)/2*lambda[4]:
      c := sqrt(2)/2*lambda[3]:
      a := 1:
      A := (b^2 + c^2 - a^2)/(2*b*c);

$A:=\frac{2k^2}{\sqrt{4+4k+2k^2}\sqrt{4-4k+2k^2}}$

• The principal stretches are given by the square roots of the eigenvalues of $𝑪$. The principal directions are the eigenvectors of $𝑪$.

Once again, we use Maple for our calculations.

      CC := linalg[matrix](3,3):
      for i from 1 to 3 do
        for j from 1 to 3 do
          CC[i,j] := eval(C[i,j], k=0.4);
        end do;
      end do;
      evalm(CC);
      eigvals := eigenvals(CC);
      PrinStretch[1] := sqrt(eigvals[1]);
      PrinStretch[2] := sqrt(eigvals[2]);
      PrinStretch[3] := sqrt(eigvals[3]);
      PrinDir := eigenvects(CC);

$CC:=\left[\begin{array}{ccc}1& 0.4& 0\\ 0.4& 1.16& 0\\ 0& 0& 1\end{array}\right]$

$\text{eigvals}:=0.67,1.0,1.49$

$\text{PrinStretch}[1]:=0.82,\text{PrinStretch}[2]:=1.0,\text{PrinStretch}[3]:=1.22$

$\text{PrinDir}[1]:=[0.77,-0.63,0.];\text{PrinDir}[2]:=[0,0,1];\text{PrinDir}[3]:=[0.63,0.77,0.]$