# Introduction to 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 ${\widehat {\mathbf {e} }}_{1}$ , ${\widehat {\mathbf {e} }}_{2}$ , $(1/{\sqrt {2}})({\widehat {\mathbf {e} }}_{1}+{\widehat {\mathbf {e} }}_{2})$ , and $(1/{\sqrt {2}})({\widehat {\mathbf {e} }}_{1}-{\widehat {\mathbf {e} }}_{2})$ .
4. Find the orthogonal shear strain between the reference material directions ${\widehat {\mathbf {e} }}_{1}$ and ${\widehat {\mathbf {e} }}_{2}$ . Also between directions $(1/{\sqrt {2}})({\widehat {\mathbf {e} }}_{1}+{\widehat {\mathbf {e} }}_{2})$ and $(1/{\sqrt {2}})({\widehat {\mathbf {e} }}_{1}-{\widehat {\mathbf {e} }}_{2})$ .
5. Find principal stretches and principal directions of stretch ($\kappa =0.4\,$ ).

### Solution:

• 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 ${\widehat {\mathbf {G} }}$ are given by
$\lambda ({\widehat {\mathbf {G} }})={\sqrt {{\widehat {\mathbf {G} }}:{\boldsymbol {C}}:{\widehat {\mathbf {G} }}}}$ where ${\boldsymbol {C}}\,$ is the Cauchy-Green deformation tensor

${\boldsymbol {C}}={\boldsymbol {F}}^{T}\bullet {\boldsymbol {F}}$ We will use Maple to calculate the stretches in the four directions.

      with(linalg):
x := array(1..3): X := array(1..3):
x := X + k*X: x := X: x := X:
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 := sqrt(evalm(evalm(e1&*C)&*transpose(e1))[1,1]);
lambda := sqrt(evalm(evalm(e2&*C)&*transpose(e2))[1,1]);
lambda := simplify(sqrt(evalm(evalm(e1pe2&*C)&*transpose(e1pe2))[1,1]));
lambda := simplify(sqrt(evalm(evalm(e1me2&*C)&*transpose(e1me2))[1,1]));

$F:={\begin{bmatrix}1&k&0\\0&1&0\\0&0&1\end{bmatrix}}$ $C:={\begin{bmatrix}1&k&0\\k&k^{2}+1&0\\0&0&1\end{bmatrix}}$ $\lambda _{1}:=1\,$ $\lambda _{2}:={\sqrt {k^{2}+1}}$ $\lambda _{3}:={\frac {\sqrt {4+4k+2k^{2}}}{2}}$ $\lambda _{4}:={\frac {\sqrt {4-4k+2k^{2}}}{2}}$ The following figure shows that the calculated stretches are correct.

• The orthogonal shear strain between two orthogonal units vectors ${\widehat {\mathbf {G} }}_{1}$ and ${\widehat {\mathbf {G} }}_{2}$ in the reference material co-ordinate system is given by
$\gamma ({\widehat {\mathbf {G} }}_{1},{\widehat {\mathbf {G} }}_{2})=\sin ^{-1}\left({\frac {{\widehat {\mathbf {G} }}_{1}:\mathbf {C} :{\widehat {\mathbf {G} }}_{2}}{\lambda ({\widehat {\mathbf {G} }}_{1})\lambda ({\widehat {\mathbf {G} }}_{2})}}\right)$ 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*lambda:
ratio1 := numer1/denom1:
gam1 := arcsin(ratio1);
numer2 := simplify(evalm(evalm(e1pe2&*C)&*transpose(e1me2))[1,1]):
denom2 := lambda*lambda:
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.

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:
c := sqrt(2)/2*lambda:
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 ${\boldsymbol {C}}$ . The principal directions are the eigenvectors of ${\boldsymbol {C}}$ .

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 := sqrt(eigvals);
PrinStretch := sqrt(eigvals);
PrinStretch := sqrt(eigvals);
PrinDir := eigenvects(CC);

$CC:={\begin{bmatrix}1&0.4&0\\0.4&1.16&0\\0&0&1\end{bmatrix}}$ ${\text{eigvals}}:=0.67,1.0,1.49$ ${\text{PrinStretch}}:=0.82~,~~{\text{PrinStretch}}:=1.0~,~~{\text{PrinStretch}}:=1.22$ ${\text{PrinDir}}:=[0.77,-0.63,0.]~;~~{\text{PrinDir}}:=[0,0,1]~;~~{\text{PrinDir}}:=[0.63,0.77,0.]$ 