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[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 := \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[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.

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 $$\boldsymbol{C}$$. 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[1] := sqrt(eigvals[1]); PrinStretch[2] := sqrt(eigvals[2]); PrinStretch[3] := sqrt(eigvals[3]); 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}[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.] $$