What Does Simple Shear Deformation Tell Us About Material Behavior?

[Bertbos]

Homework Statement

Let {i_i}, i = 1, 2, 3 be an orthonormal basis as shown in Figure 2 and consider a
simple shear deformation from R to R′ defined as
$\hat{y}$(x) = x + γ(x · i₂)i₁ , where the scalar γ corresponds to amount of shear.

Homework Equations

For this homogeneous deformation, compute the deformation gradient F
and the translation vector c. Compute the change in length of fibers of
unit length in R aligned with the basis vectors i_i (i = 1, 2, 3). What can
you say about fibers aligned with i₁ and i₃? Compute the change in angle
of pairs of fibers aligned with i_i, i_j (for i, j = 1, 2, 3, i ≠ j).

The Attempt at a Solution

If I could get $\hat{y}$(x) in a form of $\hat{y}$(x) = Fx + c I could compute the different variables, but I don't know how to get the equation in the right format. Computing F can be done by F = ∇y, with ∇=∇_x ; however, I don't know how to compute the translation vector c. Could anyone help me?

 

[Chestermiller]

Let x = x₁i₁+x₂i₂+x₃i₃

y=y₁i₁+y₂i₂+y₃i₃

So, $$x\centerdot i_2=x_2$$

and γ(x · i2)i1=γx₂i₁

so
y₁i₁+y₂i₂+y₃i₃
=x₁i₁+x₂i₂+x₃i₃
+γx₂i₁
So now, resolve this equation into its 3 components (i.e., into the coefficients of the three unit vectors). Also, write out y = Fx + c in component form, and compare the expressions.

 

[Bertbos]

Let x = x₁i₁+x₂i₂+x₃i₃

y=y₁i₁+y₂i₂+y₃i₃

So, $$x\centerdot i_2=x_2$$

and γ(x · i2)i1=γx₂i₁

so
y₁i₁+y₂i₂+y₃i₃
=x₁i₁+x₂i₂+x₃i₃
+γx₂i₁
So now, resolve this equation into its 3 components (i.e., into the coefficients of the three unit vectors). Also, write out y = Fx + c in component form, and compare the expressions.

Thank you very much. Sometimes I've trouble to write the equations is the right form. Your explanation was very clear and now it's easy to me to solve the problem!

 

[Chestermiller]

Thank you very much. Sometimes I've trouble to write the equations is the right form. Your explanation was very clear and now it's easy to me to solve the problem!

There's an even simpler way of doing it that might appeal to you even more.

Write $$x=(i_1i_1+i_2i_2+i_3i_3)\centerdot x$$
and $$(x \centerdot i_2)i_1=i_1i_2\centerdot x$$

So $$y=x+γ(x \centerdot i_2)i_1=(i_1i_1+i_2i_2+i_3i_3)\centerdot x+γi_1i_2\centerdot x=(i_1i_1+γi_1i_2+i_2i_2+i_3i_3)\centerdot x$$

So, $F=i_1i_1+γi_1i_2+i_2i_2+i_3i_3$