# Introduction to Elasticity/Kinematics example 1

## Example 1

Take a unit cube of material. Rotate it 90 degrees in the clockwise direction around the z-axis. Calculate the strains. Discuss your results - their accuracy and the reasons for your conclusions.

### Solution

The strains are related to displacements by

$\epsilon _{xx}={\frac {\partial u}{\partial x}};~\epsilon _{yy}={\frac {\partial v}{\partial y}};~\epsilon _{zz}={\frac {\partial w}{\partial z}};~\gamma _{xy}={\frac {\partial u}{\partial y}}+{\frac {\partial v}{\partial x}};~\gamma _{yz}={\frac {\partial v}{\partial z}}+{\frac {\partial w}{\partial y}};~\gamma _{zx}={\frac {\partial w}{\partial x}}+{\frac {\partial u}{\partial z}}$ Let us consider rotation about the center of the cube. Since the problem concerns a pure rotation, a cylindrical co-ordinate system is appropriate. This problem also provides us a easy way of trying out Maple. Here are the steps that you can follow to find the strains at a point in the cube.

 r := sqrt(x^2+y^2); 

$r:={\sqrt {x^{2}+y^{2}}}$ theta := arctan(y/x); 

$\theta :=arctan({\frac {y}{x}})$ x1 := r*cos(theta); 

$x1:={\frac {\sqrt {x^{2}+y^{2}}}{\sqrt {1+{\frac {y^{2}}{x^{2}}}}}}$ y1 := r*sin(theta); 

$y1:={\frac {{\sqrt {x^{2}+y^{2}}}\,y}{x\,{\sqrt {1+{\frac {y^{2}}{x^{2}}}}}}}$ x2 := r*cos(theta+Pi/2); 

$x2:=-{\frac {{\sqrt {x^{2}+y^{2}}}\,y}{x\,{\sqrt {1+{\frac {y^{2}}{x^{2}}}}}}}$ y2 := r*sin(theta+Pi/2); 

$y2:={\frac {\sqrt {x^{2}+y^{2}}}{\sqrt {1+{\frac {y^{2}}{x^{2}}}}}}$ u := x2 - x1; 

$u:=-{\frac {{\sqrt {x^{2}+y^{2}}}\,y}{x\,{\sqrt {1+{\frac {y^{2}}{x^{2}}}}}}}-{\frac {\sqrt {x^{2}+y^{2}}}{\sqrt {1+{\frac {y^{2}}{x^{2}}}}}}$ v := y2 - y1; 

$v:={\frac {\sqrt {x^{2}+y^{2}}}{\sqrt {1+{\frac {y^{2}}{x^{2}}}}}}-{\frac {{\sqrt {x^{2}+y^{2}}}\,y}{x\,{\sqrt {1+{\frac {y^{2}}{x^{2}}}}}}}$ epsx := simplify(diff(u,x)); 

$epsx:=-{\frac {\sqrt {x^{2}+y^{2}}}{x\,{\sqrt {\frac {x^{2}+y^{2}}{x^{2}}}}}}$ epsy := simplify(diff(v,y)); 

$epsy:=-{\frac {\sqrt {x^{2}+y^{2}}}{x\,{\sqrt {\frac {x^{2}+y^{2}}{x^{2}}}}}}$ gamxy := simplify(diff(u,y) + diff(v,x)); 

$gamxy:=0$ From the above Maple calculation, and noting that there is no motion in the $z$ direction, the strains in the cube are

$\epsilon _{xx}=-1;~\epsilon _{yy}=-1;\epsilon _{zz}=0;\gamma _{xy}=0;\gamma _{yz}=0;\gamma _{zx}=0$ A pure rigid body rotation should not result in any non-zero strains.

Therefore, the measure of strain we have used is not appropriate for large rigid body motions.