# Rigid Body Rotations and Engineering Strain

Problem

Consider the following 2D rigid body in the $x$-$y$-plane, which is rigidly rotated around the $z$-axis with the angle $\beta$: (Image Source: Continuum Mechanics - Small Scale Strains - Limitations of Small Strain Equations - Effect of Rotations on Strains)

A point $[X, Y]^T$ will then be rotated to a new point $$[x, y]^T = \ldots$$ $$\ldots = [X \cdot \cos(\beta) - Y \cdot \sin(\beta), \ldots$$ $$\ldots X \cdot \sin(\beta) + Y \cdot \cos(\beta)]^T$$

The engineering strains for this rigid body rotations can thus be given as $$\epsilon_{xx} = \cos(\beta)-1$$ $$\epsilon_{yy} = \cos(\beta)-1$$ $$\epsilon_{xy} = 0$$

This means that normal strains will be induced for (large) rigid body deformations. For $\beta=90°$, they will be $\epsilon_{xx} = \epsilon_{yy} = -1$. For $\beta=180°$, they will be $\epsilon_{xx} = \epsilon_{yy} = -2$.

In a text book, I found a corresponding picture, which shows a FEM simulation based on engineering strains. As a result of the rigid body rotation, the body grew in size: This seems to correspond to a gif I found on the German Wikipedia site (here a static version) showing the von Mises stresses: Question

Why does the body grow in size instead of getting smaller? The normal strains are always smaller than zero, which means that the stresses are also smaller than zero (for a linear stress-strain-relationship). Would that not entail compression?
What do I miss?

• The pictures from the book show a 90° rotation. And why do you think that rotating a body does lead to a change of its shape? It is not a dynamic rotation but can be viewed as a static rotation. – MrYouMath Apr 23 '17 at 21:04
• @MrYouMath This is related to continuum mechanics and different definitions of strain, especially the effects of using engineering strain for large rigid body rotations, whis one must never do. – Discbrake Apr 24 '17 at 6:16

The question seems to be "why does the FEM seemingly produce results that are contradictory to theory?"

Consider a square with nodes

id  x    y
1 -1.0 -1.0
2  1.0 -1.0
3  1.0  1.0
4 -1.0  1.0


The FE displacement field is $$u_x(x,y) = \sum_{j=1}^4 u_x^j N_j(x,y) ~,~~ u_y(x,y) = \sum_{j=1}^4 u_y^j N_j(x,y)$$

The strain field is $$\varepsilon_{xx}(x,y) = \sum_{j=1}^4 u_x^j \frac{\partial N_j(x,y)}{\partial x} ~,~~ \varepsilon_{yy}(x,y) = \sum_{j=1}^4 u_y^j \frac{\partial N_j(x,y)}{\partial y}$$ and $$\varepsilon_{xy}(x,y) = \tfrac{1}{2}\sum_{j=1}^4 \left[u_x^j \frac{\partial N_j(x,y)}{\partial y} + u_y^j \frac{\partial N_j(x,y)}{\partial x}\right]$$ The nodal shape functions are $$N_1(x,y) = \frac{(1-x)(1-y)}{4} ~,~~ N_2(x,y) = \frac{(1+x)(1-y)}{4}$$ $$N_3(x,y) = \frac{(1+x)(1+y)}{4} ~,~~ N_4(x,y) = \frac{(1-x)(1+y)}{4}$$ The gradients of the shape functions are \begin{align} G_{1x} := \frac{\partial N_1(x,y)}{\partial x} & = -\frac{(1-y)}{4} ~,~~ G_{1y} := \frac{\partial N_1(x,y)}{\partial y} = -\frac{(1-x)}{4} \\ G_{2x} := \frac{\partial N_2(x,y)}{\partial x} & = \frac{(1-y)}{4} ~,~~ G_{2y} := \frac{\partial N_2(x,y)}{\partial y} = -\frac{(1+x)}{4} \\ G_{3x} := \frac{\partial N_3(x,y)}{\partial x} & = \frac{(1+y)}{4} ~,~~ G_{3y} := \frac{\partial N_3(x,y)}{\partial y} = \frac{(1+x)}{4} \\ G_{4x} := \frac{\partial N_4(x,y)}{\partial x} & = -\frac{(1+y)}{4} ~,~~ G_{4y} := \frac{\partial N_4(x,y)}{\partial y} = \frac{(1-x)}{4} \end{align} Therefore the strain field is $$\begin{bmatrix} \varepsilon_{xx}(x,y) \\ \varepsilon_{yy}(x,y) \\ 2\varepsilon_{xy}(x,y) \end{bmatrix} = \begin{bmatrix} G_{1x} & G_{2x} & G_{3x} & G_{4x} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & G_{1y} & G_{2y} & G_{3y} & G_{4y} \\ G_{1y} & G_{2y} & G_{3y} & G_{4y} & G_{1x} & G_{2x} & G_{3x} & G_{4x} \end{bmatrix} \mathbf{u}$$ where $$\mathbf{u} = \begin{bmatrix} u_x^1 & u_x^2 & u_x^3 & u_x^4 & u_y^1 & u_y^2 & u_y^3 & u_y^4 \end{bmatrix}^T$$ Consider the case where the square is rotated by 90 degrees clockwise around node 1 so that the new positions of the nodes are

id  x    y
1 -1.0 -1.0
2 -1.0 -3.0
3  1.0 -3.0
4  1.0 -1.0


Note that we are not assuming any deformation of the square.

Then the displacements are

id  u_x  u_y
1  0.0 0.0
2 -2.0 -2.0
3  0.0 -4.0
4  2.0 -2.0


If we plug in the values, at node 1 we get

B1 = [[G1x(1) G2x(1) G3x(1) G4x(1) 0 0 0 0];...
[0 0 0 0 G1y(1) G2y(1) G3y(1) G4y(1)];...
[G1y(1) G2y(1) G3y(1) G4y(1) G1x(1) G2x(1) G3x(1) G4x(1)]]
=   -0.50000   0.50000   0.00000  -0.00000   0.00000   0.00000 0.00000   0.00000
0.00000   0.00000   0.00000   0.00000  -0.50000  -0.00000 0.00000   0.50000
-0.50000  -0.00000   0.00000   0.50000  -0.50000   0.50000 0.00000  -0.00000


and the nodal displacement vector is

u = [0 -2 0 2 0 -2 -4 -2]


The strain at node 1 is then

eps1 = B1*u' = [-1 -1 0]'


Therefore, the finite element solution is identical to your solution and just says that stresses will develop in the element due to pure rigid body rotation even if the element does not deform.

The finite element method is typically implemented in displacement form. So you specify the displacements and then find the stresses in the element.

An alternative would be to specify moments that would rotate the element. That's tricky to do. I'm more inclined to believe that the figure in the textbook is just meant to be an illustration.

Engineering strains are only valid mathematically for infinitesimally small rotations.

If you want to work with finite rotations (of arbitrary size) you need to use a different strain measure, such as Green strain.

• not infitesimaly just small, as per small angles simplification. So somewhat accurate to a degree though certainly not valid at 15 degrees for example. But yes what you say is true. – joojaa Apr 24 '17 at 17:38
• Yes, I know that. My question is about the induced effect of using engineering strain (which is of course wrong, but important for the understanding) - namely why the body will appear to grow in size instead of shrinking. eps_xx and eps_yy can be calculated to be always smaller than zero, yet the body grows according to FEM simulations. Why? – Discbrake Apr 24 '17 at 18:14
• @user5564832 "Why?" Wrong is wrong. Garbage in, garbage out. There's nothing else much to say about it. The reason it "appears to grow" is lines that are tangent to a circle lie entirely outside the circle. The eps_xx and eps_yy you are calculating are also nonsense, whether they are positive or negative. – alephzero Apr 27 '17 at 14:18
• alephzero, I strongly disagree. In my opinion, one needs to know all possible errors of numerical methods and their effects in order to fully understand them. Sure it is possible to click on some buttons in ANSYS without fundamental knowledge and generate colorful pictures, but that is not always the best way. And no, the eps_xx and eps_yy above are not nonsense. – Discbrake Apr 28 '17 at 18:01