Consider a unit square body as shown in the figure below. The body is subjected to the deformation field \(u = -ay\) and \(v = ax\), where 'a' is a constant. Due to the application of this deformation field, the body undergoes ________ in the x-y plane.
Show Hint
Compute the strain tensor from the displacement field first; if every strain component is zero but the antisymmetric rotation term is not, the motion is a pure rotation.
Step 1: Recall the exact form of a small rotation.
Rotating a point $(x,y)$ by a small angle $\theta$ (in radians) about the origin sends it to
\[
x' = x\cos\theta - y\sin\theta \approx x - \theta y, \qquad y' = x\sin\theta + y\cos\theta \approx \theta x + y
\]
using $\sin\theta\approx\theta$, $\cos\theta\approx1$ for small $\theta$. So the displacement of that point is
\[
u = x'-x \approx -\theta y, \qquad v = y'-y \approx \theta x
\]
Step 2: Compare this template directly with the given field.
The question gives $u=-ay$ and $v=ax$. Matching term by term with $u=-\theta y$, $v=\theta x$ shows $\theta=a$ exactly. The given field is nothing but a small rotation of angle $a$.
Step 3: Confirm by moving the actual corners of the unit square.
Take the corner $(1,0)$: it moves to $(1-a\cdot0,\ a\cdot1+0)=(1,a)$.
Take the corner $(1,1)$: it moves to $(1-a,\ a+1)$.
Take the corner $(0,1)$: it moves to $(-a,1)$.
The distance between adjacent corners, for example from $(1,0)$ to $(1,1)$ before deformation (length 1) and from $(1,a)$ to $(1-a,\,a+1)$ after deformation, is
\[
\sqrt{(1-(1-a))^2+((a+1)-a)^2} = \sqrt{a^2+1}
\]
For a genuinely small rotation angle $a$ this is 1 to first order (the $a^2$ term is negligible), so every side keeps its original length and every corner keeps its right angle. Nothing stretches and nothing shears; the square simply spins about the origin.
Step 4: Conclude.
The whole shape moves as one rigid piece, changing orientation only, which is the definition of rigid body rotation, not biaxial deformation, shear, or bending.
\[
\boxed{\text{Rigid body rotation}}
\]
