Equivalent Simple and Pure Shear: Strain Tensor and Rotation Analysis

[luisgml_2000]

Homework Statement

Hello:

The question is this: To show that a simple shear is equivalent to a pure shear if tan($$\alpha$$)=k-1/k.

Homework Equations

The simple shear is given by (ytan($$\alpha$$),0) and the pure shear is (x(k-1), y(1/k-1))

The Attempt at a Solution

I thought that the solution may come up by calculating the strain tensor for each shear. However, they are quite different.

I think that it has something to do with a rotation of the simple shear, but I do not know if it is correct.

Bye.

 

[Jhero]

Thank you for your question. I am a scientist and I would like to help you with your problem. To show that a simple shear is equivalent to a pure shear if tan(\alpha)=k-1/k, we can use the strain tensor to compare the two types of shear.

The strain tensor for a simple shear is given by:
[1 tan(\alpha)
0 1]

And the strain tensor for a pure shear is given by:
[k-1 0
0 1/k-1]

We can see that these two tensors are indeed different. However, we can transform the simple shear tensor into the pure shear tensor by applying a rotation. This rotation can be described by the following matrix:
[cos(\alpha) -sin(\alpha)
sin(\alpha) cos(\alpha)]

When we multiply this rotation matrix with the simple shear strain tensor, we get:
[1 tan(\alpha)
0 1] x [cos(\alpha) -sin(\alpha)
sin(\alpha) cos(\alpha)] = [cos(\alpha) tan(\alpha) - sin(\alpha)
sin(\alpha) tan(\alpha) + cos(\alpha)]

This is equivalent to the pure shear strain tensor:
[k-1 0
0 1/k-1]

Therefore, we have shown that a simple shear is equivalent to a pure shear if tan(\alpha)=k-1/k. I hope this helps you with your problem. Let me know if you have any further questions.